A cuboid water tank is partially filled with 1 6 0 c m 3 of water. Depending on which face it rests on, the height of water in the tank is 2 c m , 5 c m , or 4 c m .
What is the volume of this tank ( in c m 3 ) ?
Note: The diagram is not drawn to scale.
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You have a typo, l 2 w 2 h 2 = 1 0 2 4 0 0 , not 1 2 4 0 0 0 .
After look at the answer, I think the figure is actually drawn to scale.
I read the depths as 2,3,5. How stupid can you get? Ed Gray
The answer is.....indeterminable.
There is no correlation between the depth of the water in the tank and the overall height of the tank.
The key is the phrase “the diagram is not drawn to scale”.
The height might be 2 inches, 5 inches or 50 feet. The depth of the water in each case shown will remain the same as shown since the height of the water is based on the length times the width of the container holding it, (and the volume of the water) not the container’s height.
I put 10 cubic feet of water in a bathtub that is 2 x 5 feet in dimension.
How high is the water level? 1 foot.
How high are the walls of the tub? Can’t say.
Thetexan
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According to the problem, the three diagrams are what the box looks like when a different face is on the bottom. What may be the height in one tank would be the length/width in the other two tanks.
I didn't think to multiply the equations like this. I used a process of simultaneous equations. But, while it wasn't as quick as what you did, I suppose that at least I ended up having the specific dimensions of the tank as well!
I dont understand
Eloquent solution, very well done!
I ended up drawing the box in it's translations and colour coding the sides to realize the relationships of sides and to determine Areas. Then with Areas 32, 40 and 80 I tried to remember the relationship of Area to volume but then looked at the answer above to realize more math was needed to build a solution out from the Squares to then solve for V. Thanks for the simple solution.
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2x2=4, 5x2=10, 4x2=8, 4x10x8=320
If the cuboid's dimensions are a , b and c , then, the given information can be summarized in the following equations: 2 a b = 1 6 0 5 b c = 1 6 0 4 a c = 1 6 0 Multiplying them we find: ( a b c ) 2 = V 2 = 4 0 1 6 0 2 ⋅ 1 6 0 = ( 1 6 0 ⋅ 2 ) 2 ⇒ V = 1 6 0 ⋅ 2 = 3 2 0
I believe it should be V^2 = (160.2)^2, not (160.4)^2. For the rest the proof is perfect.
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I second his correction
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Let the dimensions of the tank be x 1 , x 2 , x 3 and the heights of the water y 1 , y 2 , y 3 . Now x 1 y 1 = x 1 x 2 x 3 y 1 x 2 x 3 = V t V w = : α , etc. , where V w is the volume of the water and V t is the volume of the tank. Therefore y 1 y 2 y 3 = α 3 x 1 x 2 x 3 = α 3 V t = α 2 V w . Thus α = V w y 1 y 2 y 3 = 1 6 0 2 ⋅ 5 ⋅ 4 = 2 1 , showing that the tank is half-full; to finish the problem, V t = α V w = 1 / 2 1 6 0 = 3 2 0 .
what is alpha
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The fraction of the volume that is filled. See the first line of equations, where I wrote: ⋯ = V t V w = : α . (The notation : = α is shorthand for: "Call this number α .")
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Note: The diagrams are not drawn to scale.
Consider my diagram. Let the side lengths of the rectangular prism be a , b and c . The volume of the cuboid is a b c . Since the volume of water is constant we have the three equations: 1 6 0 = 4 a c ( 1 ) , 1 6 0 = 5 b c ( 2 ) and 1 6 0 = 2 a b ( 3 )
Multiplying the three equations, we get
1 6 0 ( 1 6 0 ) ( 1 6 0 ) = 4 a c ( 5 b c ) ( 2 a b )
4 0 9 6 0 0 0 = 4 0 a 2 b 2 c 2
Dividing both sides by 4 0 , we get
1 0 2 4 0 0 = a 2 b 2 c 2
Extracting the square root of both sides, we get
3 2 0 = a b c
Nice. Can you explain the magic part "Multiplying the three equations"?
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What we want is the value of a b c . If we multiply the three equations, we can get the value of a 2 b 2 c 2 , from here we can get the value of a b c . The longer solution is by substitution. There are three equations and three unknowns. We find that a = 1 0 , b = 8 and c = 4 .
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I looked at the pictures and noticed the volume of water was half of the capacity. Chose the quick way, and doubled each volume, then multiplied the capacities of L x W x H in my head. Thanks to everyone who did it long hand. If there had been no pictures I’d have been reliant on a pen & paper
The picture is defined as 'not to scale' though, so you shouldn't rely on that bit of information.
Considering the dimensions of the solid, a, b and c. The area of each face is: a ∗ b = 1 6 0 / 2 = 8 0 b ∗ c = 1 6 0 / 5 = 3 2 and a ∗ c = 1 6 0 / 4 = 4 0 .
If we make ( a ∗ b ) / ( a ∗ c ) = 8 0 / 4 0 , we have b = 2 ∗ c , then replacing c in the remaining equation, c ∗ 2 c = 3 2 , so c = 4 . Then, b = 2 ∗ c = 8 , and a ∗ 4 = 1 0 , a = 1 0 .
The solid volume is a ∗ b ∗ c = 1 0 ∗ 8 ∗ 4 = 3 2 0 .
If the volume is 160 cm^3 and the heights are 2, 4, and 5 cm, I'm looking at areas of 32, 40 and 80 cm^2. These are RECTANGLES so suddenly I'm thinking 4, 8, and 10 as the sides of a box. I hadn't even found a pencil, but 4, 8, and 10 start screaming '320!!' so I clicked it. If you can smell the answer, why engage your brain?
Your method is exactly the same as mine. As an aside the four listed options were poorly chosen, since only one of them could be arrived at by multiplying the numbers 2, 4 and 5 and no other numbers - had all four options given been multiples of these numbers a little more thought would have been required, although ti would still have been easy.
I'm with you, Bill. But unless that tank is transparent, after all the figuring, I'm gonna break out old Trusty Lufkin and make sure. If the volume is compromised by an unknown internal feature, no amount of pencil jiggling is going to make it fit in the closet. I looked at 160cm^3 / 2, giving the cm^2 of 80, with a ratio of 5:4 on the two remaining sides, the answer seemed to leap off the page.
Can you explain it with more details?
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You mean like 'unknown volume divided by known depth is unknown area?' There's no other details V = 160, h = 2,4,5 -> areas 80, 40, 32. So box dimensions of 4, 8, 10 POPPED INTO MY HEAD AS DID THE PRODUCT, 320. So sorry, no new details, that was the full process.
Let the sides of the box be x,y,z (in cm) So from the question, Volume of water = 160cm^3 = 2xy = 5xy = 4yz = 160. So, We get 3 equations, xy=80, xz=32, yz=40. => 80/y = 32/z => y=80.z/32 => 80.z^2/32 = 40 => z^2 = 16 .'. z=4 (as side length can't be a negative number)
So, volume = x.y.z = (80).4 = 320 cm^3
Look at the first picture. We know that the total volume of water is 160 and that the height is 2, so therefore, the other 2 sides, when multiplied, must give 80.
So now we know that 80 multiplied by a certain number must give us the volume.
Finally, we can deduce the total volume is 320 because that is the only multiple of 80.
This is the solution I was looking for!! Thank you:) I knew there was a way to do it without involving calculation of all side lengths!
Now that I think of it, you can actually do that with any/all 3 sides. 320 is the only multiple of 40, 80, and 32
Except that no where is it stated that the sides are integers.
An elegant solution that draws an additional piece of information from the multiple choice format of the question!
Simple answer, 160 × 2 =320,thats the fastest way to the correct answer if your a Genius like me..
In logical view:: Cuboid partially(half) tank filled with water V/2 = 160 in any heights, so heights apart ,then fully filled capacity V = 160*2
I thought this is so many times more complicated. So I was complivating and complicating but at the end it was just 160•2=320 or 4•8,94427191^2=320
Another approach, shorter and faster: without pen and paper. Tank volume V is x times 160. V = 160x = (2x) (4x) (5x) as each side length of the tank is x times the water level at any of the three situations shown. Well, calculate in mind gives x=2 and V=320.
Btw,water level is perfectly scaled in the "non scaled" picture.
I guessed 320 as it is exactly twice 160.
The problem would have been better if the heights were (say) 2, 3, 6 and the possible answers rounded to nearest integer. (INT(337.31) ).
Given that most people could not do the problem without pen and paper anyway, answers which were "not neat" would be better.
The diagram was drawn to scale.
It is a trick question to make engineers do busy work. The second word and the first height is all you need to know, and it's value is arbitrary to the solution. The classification of cuboid implies all opposing sides are equal. Therefore, if it is 'X'cm on one side and you turn it over 180 degrees in any direction the cuboid requires that the height remain the same enforcing a solution of double the volume.
Take sides of cubes as 2,3,5 and take its largest multiple that is 320
Rather intuitively looking at the first figure, if you stack two such volumes of water it will equal the volume of the cuboid. It is already given one volume of such water is 160 cubic centimetre. So twice of that will be 320 cubic centimetre.
Based on the diagram, the tank looked half full to me. If the given is 160, Then double is 320.
Since the question is multiple choice, I know that one of the answers is correct. Not really how to solve the problem I guess, but how to answer the question.
Apologies to all who know what they are doing. : ) Cheers.
You can have a 3 system equation and solve by substitution:
2AB=160
5BC=160
4AC=160
Taking the 3rd equation A=(160/4C), you get:
A=40/C
Now substitute on 1st equation 2B(40/C)=160, you get:
B=2C
Then substitute in 2nd equation 5(2C)(C)=160. C=4
Now we get B in same equation 4(5)B=160, B=8
From there in 3rd equation, 4(4)A=160. A=10
ABC=320
Let's call the different dimensions of the box x,y and z.Then we have the volume of the water to be 2 x y, 4 y z and 5 x z, and x y=80, y z=40 and x*z=32. Multiplying the equations gives x 2 y 2 z 2 = 1 0 2 4 0 0 and xyz = 320
If the level was 1 cm, then looking at the 2cm tank it would be full. So 1 cm x 2 = 320. bold text
I did the problem as stated, then I generalized it.
Below is the generalization:
Problem:
Let a , b , c and m be positive real numbers.
A cuboid water tank is partially filled with m c m 3 of water. Depending on which face it rests on, the height of water in the tank is a c m , b c m , or c c m .
What is the volume of this tank ( in c m 3 ) ?
a x y = m
b x z = m
c y z = m
⟹ a b c x 2 y 2 z 2 = m 3 ⟹ x 2 y 2 z 2 = a b c m 3 ⟹ x y z = a b c m 3
∴ The volume V = a b c m 3
Using m = 1 6 0 , a = 2 , b = 5 and c = 4 ⟹ V = 4 0 1 6 0 ∗ 1 6 0 2 = 2 ∗ 1 6 0 = 3 2 0 .
I feel so stupid looking at the very mathematical solutions here, I just looked at the diagram even though it wasn't to scale, it looked as of the water was about half of what the diagram was, so I just doubled the volume of the water.
We can easily calculate the side lengths of the water tank doubling provided hights of water accordingly. It will give us 4, 10 and 8 cm. Now we know how to calculate the tank volume: 4 × 10 × 8 = 3 2 0
I don't think that's a good solution. In this case, it may works, but the problem does not state that the water appears to fill half the tank. In exemple, on the first diagram, you can't say the side that doesn't fill up completely is equal to 4 because the water reaches height 2.
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Call the dimensions of the box l , w and h . Then the volumes of the water in the three boxes are 2 ⋅ l ⋅ w = 5 ⋅ w ⋅ h = 4 ⋅ l ⋅ h = 1 6 0 , so l w = 8 0 , w h = 3 2 and l h = 4 0 . Multiplying these three equations together gives us l 2 w 2 h 2 = 1 0 2 4 0 0 and so V = l w h = 3 2 0